$\lim_{x\to -2}\dfrac{\sqrt{3x+10}-2}{x+2}=$
Substituting $x=-2$ into $\dfrac{\sqrt{3x+10}-2}{x+2}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{\sqrt{3x+10}-2}{x+2} \\\\ &=\dfrac{\sqrt{3x+10}-2}{x+2}\cdot\dfrac{\sqrt{3x+10}+2}{\sqrt{3x+10}+2} \gray{\text{Rationalize the numerator}} \\\\ &=\dfrac{(3x+10)-2^2}{(x+2)(\sqrt{3x+10}+2)} \\\\ &=\dfrac{3\cancel{(x+2)}}{\cancel{(x+2)}(\sqrt{3x+10}+2)} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{3}{\sqrt{3x+10}+2} \text{, for }x\neq -2 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-2$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\sqrt{3x+10}-2}{x+2}=\dfrac{3}{\sqrt{3x+10}+2}$ for all $x$ -values in the interval $(-2.5,-1.5)$ except for $x=-2$. Therefore, $\lim_{x\to -2}\dfrac{\sqrt{3x+10}-2}{x+2}=\lim_{x\to -2}\dfrac{3}{\sqrt{3x+10}+2}=\dfrac{3}{4}$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -2}\dfrac{\sqrt{3x+10}-2}{x+2}=\dfrac{3}{4}$.